My goal is to prove that:

$\displaystyle\sum\limits_{n=1}^\infty \frac{1}{n^{x}} \rightarrow \infty$ for all $ x\rightarrow 1^+$

In order to show this statement I show that no matter how big you choose $N\in \mathbb{R}$, you can always find a $\delta >0$ so $\displaystyle\sum\limits_{n=1}^\infty \frac{1}{n^{x}} = 1+\frac{1}{2^{x}}+\cdots+0>N$ when $x\in \left]1,1+\delta\right[$

My idea is to use some kind of "induction".

First/the start : N = 1.25

If we look at the case where $N = 1.25$ I conclude that i can just choose $\delta = 1$ so the following statement is true:

$\displaystyle\sum\limits_{n=1}^\infty \frac{1}{n^{x}} = 1+\frac{1}{2^{x}}+\cdots+0> 1.25$ for all $x\in \left]1,2\right[$

The "induction"-step

I now assume that the following statement is true for an $N$:

There exists a $\delta>0$ so

$\displaystyle\sum\limits_{n=1}^\infty \frac{1}{n^{x}} = 1+\frac{1}{2^{x}}+\cdots+0>N$ for all $x\in \left]1,1+\delta\right[$

Now I multiply the above equation with $x$ ($x$ is greater than $1$)

$\displaystyle x+\frac{x}{2^{x}}+\frac{x}{3^{x}}+\cdots +0>xN$ for all $x\in \left]1,1+\delta\right[$

Then I do some magic and get from the above:

$\displaystyle 1+\frac{x}{2^{x}}+\frac{x}{3^{x}}+\cdots+0>xN-x+1$ for all $x\in \left]1,1+\delta\right[$

I now see that following is true:

$xN-x+1 > N$

Because I dont need to go below $N=1.25$ and can therefore do the following calculations:

$xN-x+1 = x(N-1)+1> N \Leftrightarrow \\ x(N-1)>N-1 \Leftrightarrow \\ x>1$

From all this I can finally conclude that:

$\displaystyle 1+\frac{x}{2^{x}}+\frac{x}{3^{x}}+\cdots+0>xN-x+1>N$ for all $x\in \left]1,1+\delta\right[$

I now write the above as:

$1+\dfrac{1}{2^{x-\frac{\ln(x)}{\ln(2)}}}+\dfrac{1}{3^{x-\frac{\ln(x)}{\ln(3)}}}+\cdots+0>xN-x+1>N$ for all $x\in \left]1,1+\delta\right[$

Because I can use the following true statement:

$\dfrac{1}{x} = a^{-\frac{\ln(x)}{\ln(a)}}$

From the above I can get a new statement:

$1+\dfrac{1}{2^{x-\frac{\ln(x)}{\ln(2)}}}+\dfrac{1}{3^{x-\frac{\ln(x)}{\ln(2)}}}+\cdots+0>xN-x+1>N$ for all $x\in \left]1,1+\delta\right[$

I substitue $y = x-\frac{\ln(x)}{\ln(2)}$ and get:


This is true for all $y\in \left]1,1+\delta_{1}\right[$ where $\delta_{1}>0$ It is true, because:

for $1+\delta>x>1$ I can get $2x-2>\ln(x) \Leftrightarrow x-1>\dfrac{\ln(x)}{2} \implies y=x-\dfrac{\ln(x)}{2}>1$ and because $x$ is limited by $1+\delta$ I can find a new smaller $\delta_{2}>0$ so

$1+\dfrac{1}{2^{y}}+\dfrac{1}{3^{y}}+\cdots+0>xN-x+1>N$ is true for all $y\in \left]1,1+\delta_{2}\right[$


Now I'm done. If I choose a $N\in \mathbb{R}$ I can go iteratively from 1.25 to a value above $N$.

My question

My question is very simple. I know it's long (and "ugly"), but does it looks right? The strategy is "homemade", so I'm a bit insecure.

  • $\begingroup$ I don't see where in your induction step you go from $N$ to $N+1$ or something like that. $\endgroup$ – Rahul May 12 '12 at 21:58
  • $\begingroup$ That's why I wrote some kind of "induction" :). So it isn't induction, but the idea is that you go from a start iteratively to a high enough value. $\endgroup$ – bemyguest May 12 '12 at 22:03
  • $\begingroup$ bemyguest, Rahul's point is that you do not iterate anywhere. You have only shown that $1 + 2^{-y} + 3^{-y} + \cdots > 1.25$. You cannot even deduce from your argument that $1 + 2^{-y} + 3^{-y} + \cdots > 1.26$. $\endgroup$ – Antonio Vargas May 12 '12 at 22:05
  • $\begingroup$ I see. But I think I can fix it. $\endgroup$ – bemyguest May 12 '12 at 22:13
  • $\begingroup$ I'dont know if it helped, but I added "xN-x+1>N" a few places. I know claim that if the statement is true for N=1.25, I can then deduce that it is true for a value xN-x+1>N and therefore iterate. $\endgroup$ – bemyguest May 12 '12 at 22:21

I want to address some trouble spots in your argument. This would be too long for the comments so I'll write it here.

1) Writing "$\cdots+0$" at the end of your infinite sums is incorrect. Just write "$+\cdots$", as in

$$ 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots > 1.25. $$

Note that it is customary to write at least three full terms before the $"\cdots"$.

2) It is true that $xN - x + 1 > N$ when $N>1$ and $x>1$, but

$$ \inf_{x>1} (xN-x+1) = N, $$

so you have not made any progress; you may still only conclude that

$$ 1+\dfrac{1}{2^{x-\frac{ln(x)}{ln(2)}}}+\frac{1}{3^{x-\frac{ln(x)}{ln(3)}}}+ \cdots>N $$

for all $x \in ]1,1+\delta[$. In order for your argument to work, you would need a new lower bound whose infimum over the interval in question is strictly larger than your previous lower bound by some amount which is "large enough" so that when you repeat the process an infinite number of times the sequence of new lower bounds diverges to $\infty$.

3) Your choice $y = x-\ln x/\ln 2$ is actually less than $1$ for all $x \in ]1,2[$. At least it is less than $1$ near $x=1$, since $\frac{d}{dx}(x - \ln x / \ln 2) = 1-1/(x \ln 2) < 0$ for $x \in [1,1/\ln 2[$. Here's a plot from WolframAlpha:

enter image description here

In other words, the statement "$y \in ]1,1+\delta_2[$ for some $\delta_2 > 0$" is false, and you may not apply the argument again to these new values.

Please let me know if I could clarify anything further.

  • $\begingroup$ Thank you. I guess I have to begin from scratch. $\endgroup$ – bemyguest May 12 '12 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.